plzzz solved 6.
tomorrow my exam
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angle C = angle A = 75°
( opposite side of parallelogram are equal)
In triangle BDC
angle DBC + angle DCA + angle CDB=180°
60°+75°+angle CDB = 180°
angle CDB = 180°-135° = 45°
( opposite side of parallelogram are equal)
In triangle BDC
angle DBC + angle DCA + angle CDB=180°
60°+75°+angle CDB = 180°
angle CDB = 180°-135° = 45°
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